Question

•10 kg of water at 100 C is poured into 10 kg of water at 20 C. What is the overall entropy change if the mixing is adiabatic? Specific heat of water is: 4.2 kJ/kg-K

Answer 1

Heat released by hot water = Heat absorbed by cold water

Mass of hot water x Cp x (100 - T) = mass of cold water X Cp x (T - 20)

10 kg x 4.2 kJ/kg-K x (100 - T) = 10 kg x 4.2 kJ/kg-K x (T - 20)

100 - T = T - 20

2T = 120

T = 60 °C

Entropy change for hot water

Sh = mass x Cp x ln (T/100)

= 10 kg x 4.2 kJ/kg-K x ln (273+60)/(273+100)

= - 4.76 kJ/K

Entropy change for cold water

Sc = mass x Cp x ln (T/20)

= 10 kg x 4.2 kJ/kg-K x ln (60+273/20+273)

= 5.37 kJ/K

Overall entropy change

S = Sh + Sc

= - 4.76 + 5.37

= 0.61 kJ/kg-K